CODEWITHSUNDEEP
jqueryajaxpost
Vikram Anand Bhushan
Vikram Anand Bhushan

Reputation: 4896

Using Jquery POST() method not working properly

I have a form something like this 

    <form id="logReg" action="<?php echo $url?>" method="post">
    <input name="im_user" type="text" class="valores" id="im_user" placeholder="Email" style="height:35px; font-size:16px;" maxlength="255" />
    <input name="im_password" type="password" class="valores" placeholder="Contrase&ntilde;a" style="height:35px; font-size:16px;" id="im_password" />
    <input name="inputAds" type="submit" value="Ingresar" />
</form>

Its a login form . Then I am using Post() method of Jquery to send the login details to the server everything working fine . But the problem is in the console when i debug the response of the post method . I see the my password and id is visible.

Ajax script is 

$("#logReg").submit(function (event) {
    // Stop form from submitting normally
    event.preventDefault();
     console.log("request being sent");
    // Get some values from elements on the page:
    var $form = $(this),
        im_user = $form.find("input[name='im_user']").val(),
        im_password = $form.find("input[name='im_password']").val(),
        url = $form.attr("action");
    // Send the data using post
    var posting = $.post(url, {
        im_user: im_user,
        im_password: im_password
    });
    //Put the results in a div
    posting.done(function (data) {
        var content = $(data).find("#content");
        $("#logRegResult").empty().append(content);
    });

});

I am very new to Ajax method . Can any one explain me how can I hide the password and id from displaying in the console .

Thanks

Upvotes: 0

Views: 304

Answers (1)

Jeffpowrs
Jeffpowrs

Reputation: 4560

When you assign some value to a variable, the left hand code is executed, then that final value is assigned to the variable.

In this case it's the jQuery post object which has already sent a request, with out a specified callback. You should specify the callback like so.

var posting = $.post(url, {
    im_user: im_user,
    im_password: im_password
}).done(function (data) {
    var content = $(data).find("#content");
    $("#logRegResult").html(content);
});

* Note: as others have mentioned, this is not a secure method for sending your passwords.

Also, it seems a little odd to me to be looking for an element by id with-in the return data. Make sure that part is actually functioning. You probably want the html to be properly formatted before it's sent back from the server.

Upvotes: 1

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